18_ch 05 Mechanical Design budynas_SM_ch05

# 18_ch 05 Mechanical Design budynas_SM_ch05 -...

This preview shows page 1. Sign up to view the full content.

132 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Here r 2 o = 25, r 2 i = 9, and so σ t ω 2 = 0 . 260 386 ± 3 . 211 8 ²± 50 + 9 1 . 633(9) 3 . 211 ² = 0 . 0147 Since σ r is of the same sign, we use M2M failure criteria in the ﬁrst quadrant. From Table A-24, S ut = 31 kpsi, thus, ω = ± 31 000 0 . 0147 ² 1 / 2 = 1452 rad/s rpm = 60 ω/ (2 π ) = 60(1452) / π ) = 13 866 rev/min Using the grade number of 30 for S = 30 000 kpsi gives a bursting speed of 13640 rev/min. 5-23 T C = (360 27)(3) = 1000 lbf · in, T B = (300 50)(4) = 1000 lbf · in In xy plane, M B = 223(8) = 1784 lbf · in and M C = 127(6) = 762 lbf · in . In the xz plane, M B = 848 lbf · in and M C = 1686 lbf · in. The resultants are M B = [(1784) 2 + (848) 2 ] 1 / 2 = 1975 lbf
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online